# The Universe of Discourse

Wed, 21 Mar 2018

A couple of years ago I was reading Wikipedia's article about the the 1943 Bengal famine, and I was startled by the following claim:

"If food is so scarce, why hasn’t Gandhi died yet?"

Winston Churchill's response to an urgent request to release food stocks for India.

It was cited, but also marked with the “not in citation” tag, which is supposed to mean that someone checked the reference and found that it did not actually support the claim.

It sounded like it might be the sort of scurrilous lie that is widely repeated but not actually supportable, so I went to follow it up. It turned out that although the quotation was not quite exact, it was not misleadingly altered, and not a scurrilous lie at all. The attributed source (Tharoor, Shashi "The Ugly Briton". Time, (29 November 2010).) claimed:

Churchill's only response to a telegram from the government in Delhi about people perishing in the famine was to ask why Gandhi hadn't died yet.

I removed the “not in citation” tag, which I felt was very misleading.

Still, I felt that anything this shocking should be as well-supported as possible. It cited Tharoor, but Tharoor could have been mistaken. So I put in some effort and dug up the original source. It is from the journal entry of Archibald Wavell, then Viceroy of India, of 5 July 1944:

This appears in the published version of Lord Wavell's journals. (Wavell, Archibald Percival. Wavell: The Viceroy's journal, p. 78. Moon, Penderel, ed. Oxford University Press, 1973.) This is the most reliable testimony one could hope for. The 1973 edition is available from the Internet Archive.

A few months later, the entire article was massively overhauled by a group of anglophiles and Churchill-rehabilitators. Having failed to remove the quotation for being uncited, and then having failed to mendaciously discredit the cited source, they removed the quotation in a typical episode of Wikipedia chicanery. In a 5,000-word article, one sentence quoting the views of the then-current British Prime Minister was deemed “undue weight”, and a failure to “fairly represent all significant viewpoints that have been published by reliable sources”.

Further reading: In Winston Churchill, Hollywood rewards a mass murderer. (Tharoor again, in last week's Washington Post.)

Tue, 20 Mar 2018

In English we can sometimes turn an adjective into a verb by suffixing “-en”. For example:

black → blacken
red → redden
white → whiten
wide → widen

But not

blue → bluen*
green → greenen*
yellow → yellowen*
long → longen*

(Note that I am only looking at -en verbs that are adjective-derived present tenses. This post is not concerned with the many -en verbs that are past participles, such as “smitten” (past participle of “smite”), “spoken” (“speak”), “molten” (“melt”), “sodden” (“seethe”), etc.)

I asked some linguist about this once and they were sure it was purely morphological, something like: black, red, and white end in stop consonants, and blue, green, and yellow don't.

Well, let's see:

Stop Blacken
Brighten
Cheapen
Darken
Embolden
Fatten
Flatten
Golden
Harden
Hearten
Heighten
Louden
Open (?)
Pinken
Quicken
Quieten
Redden
Ripen
Sharpen
Shorten
Sicken
Slacken
Smarten
Straighten
Straiten
Sweeten
Thicken
Tighten
Weaken
Whiten
Widen
Biggen
Fricative Coarsen
Deafen
Enlargen
Enliven
Fasten
Freshen
Hasten
Leaven
Lengthen
Lessen
Loosen
Moisten
Roughen
Soften
Stiffen
Strengthen
Toughen
Worsen
Largen
Smoothen
Nasal   Cleanen
Dimmen
Dumben
Finen
Greenen
Longen
Slimmen
Strongen
Thinnen
Vowel   Angrien
Bluen
Dirtien
Dryen
Grayen
Highen
Lowen
Narrowen
Noisien
Saltien
Slowen
Yellowen
Nasalized
stop
Dampen
Blunten
Glide   Betteren
Bitteren
Dullen
Faren
Greateren
Moren
Nearen
Smallen
Souren
Stalen

There are some fine points:

• “Biggen” used to exist but has fallen out of use
• Perhaps I should have ommitted “strengthen” and “hasten”, which are derived from nouns, not from adjectives
• I'm not sure whether “closen”, “hotten” and “wetten” are good or bad so I left them off
• “moisten” and “soften” might belong with the stops instead of the fricatives
• etc.

but clearly the morphological explanation wins. I'm convinced.

[ Addendum: Wiktionary discusses this suffix, distinguishing it from the etymologically distinct participial “-en”, and says “it is not currently very productive in forming new words, being mostly restricted to monosyllabic bases which end in an obstruent”. ]

Mon, 19 Mar 2018

I had a fun idea this morning. As a kid I was really interested in polar coordinates and kind of disappointed that there didn't seem to be any other coordinate systems to tinker with. But this morning I realized there were a lot.

Let !!F(c)!! be some parametrized family of curves that partition the plane, or almost all of the plane, say except for a finite number of exceptions. If you have two such families !!F_1(c)!! and !!F_2(c)!!, and if each curve in !!F_1!! intersects each curve in !!F_2!! in exactly one point (again with maybe a few exceptions) then you have a coordinate system: almost every point !!P!! lies on !!F_1(a)!! and !!F_2(b)!! for some unique choice of !!\langle a, b\rangle!!, and these are its coordinates in the !!F_1–F_2!! system.

For example, when !!F_1(c)!! is the family of lines !!x=c!! and !!F_2(c)!! is the family of lines !!y=c!! then you get ordinary Cartesian coordinates, and when !!F_1(c)!! is the family of circles !!x^2+y^2=c!! and !!F_2(c)!! is the family !!y=cx!! (plus also !!x=0!!) you get standard polar coordinates, which don't quite work because the origin is in every member of !!F_2!!, but it's the only weird exception.

But there are many other families that work. To take a particularly simple example you can pick some constant !!k!! and then take

\begin{align} F_1(c): && x & =c \\ F_2(c): && y & =kx+c. \end{align}

This is like Cartesian coordinates except the axes are skewed. I did know about this when I was a kid but I considered it not sufficiently interesting.

For a more interesting example, try

\begin{align} F_1(c): && x^2-y^2 & =c \\ F_2(c): && xy & =c \end{align}

which looks like this:

I've seen that illustration before but I don't think I thought of using it as a coordinate system. Well, okay, every pair of hyperbolas intersects in two points, not one. So it's a parametrization of the boundary of real projective space or something, fine. Still fun!

In the very nice cases (such as the hyperbolas) each pair of curves is orthogonal at their point of intersection, but that's not a requirement, as with the skew Cartesian system. I'm pretty sure that if you have one family !!F!! you can construct a dual family !!F'!! that is orthogonal to it everywhere by letting !!F'!! be the paths of gradient descent or something. I'm not sure what the orthogonality is going to be important for but I bet it's sometimes useful.

You can also mix and match families, so for example take:

\begin{align} F_1(c): && x & =c \\ F_2(c): && xy & =c \end{align}

Some examples work better than others. The !!xy=c!! hyperbolas are kind of a mess when !!c=0!!, and they don't go together with the !!x^2+y^2=c!! circles in the right way at all: each circle intersects each hyperbola in four points. But it occurs to me that as with the projective plane thingy, we don't have to let that be a problem. Take !!S!! to be the quotient space of the plane where two points are identified if their !!F_1–F_2!!-coordinates are the same and then investigate !!S!!. Or maybe go more directly and take !!S = F_1 \times F_2!! (literally the Cartesian product), and then topologize !!S!! in some reasonably natural way. Maybe just give it the product topology. I dunno, I have to think about it.

(I was a bit worried about how to draw the hyperbola picture, but I tried Google Image search for “families of orthogonal hyperbolas”, and got just what I needed. Truly, we live in an age of marvels!)

Mon, 12 Mar 2018

I've been thinking for a while that I probably ought to get around to memorizing all the prime numbers under 1,000, so that I don't have to wonder about things like 893 all the time, and last night in the car I started thinking about it again, and wondered how hard it would be. There are 25 primes under 100, so presumably fewer than 250 under 1,000, which is not excessive. But I wondered if I could get a better estimate.

The prime number theorem tells us that the number of primes less than !!n!! is !!O(\frac n{\log n})!! and I think the logarithm is a natural one, but maybe there is some constant factor in there or something, I forget and I did not want to think about it too hard because I was driving. Anyway I cannot do natural logarithms in my head.

Be we don't need to do any actual logarithms. Let's estimate the fraction of primes up to !!n!! as !!\frac 1{c\log n}!! where !!c!! is unknown and the base of the logarithm is then unimportant. The denominator scales linearly with the power of !!n!!, so the difference between the denominators for !!n=10!! and !!n=100!! is the same as the difference between the denominators for !!n=100!! and !!n=1000!!.

There are 4 primes less than 10, or !!\frac25!!, so the denominator is 2.5. And there are 25 primes less than 100, so the denominator here is 4. The difference is 1.5, so the denominator for !!n=1000!! ought to be around 5.5, and that means that about !!\frac2{11}!! of the numbers up to 1000 are prime. This yields an estimate of 182.

I found out later that the correct number is 186, so I felt pretty good about that.

[ Addendum: The correct number is 168, not 186, so I wasn't as close as I thought. ]